3 years ago

In circle T below, find the measure of angle BEC (This is actually Geometry so..)

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Answer:

276.2

Step-by-step explanation:

I just subtracted the totals of AB and DC by 360. It's been a while since I've done this though.

You might be interested in

What is the greatest common factor of 24s^3 , 12s^4 , and 18s

denpristay [2]

6s. 2 and 3 divide into all these numbers but six is the greatest number that all
Divide into and they all have one s at least.

6 0

3 years ago

Read 2 more answers

A farm stand wants to sell fruit to local grocery stores in packages with baskets of apples, worth $10 apiece, and baskets of pe

Scorpion4ik [409]

The answer is B 22 apples and 28 peaches

7 0

3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$